If $A= mathbbZ[X]/(X^n+1)$, is it true that $A/mA cong mathbbZ_m[X]/(X^n+1)$?

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Let $R= mathbbZ[X]/(X^n+1)$ for some sufficiently large $n$. For $q geq 2$, I want to show that $R/qR cong mathbbZ_q[X]/(X^n+1)$.

I've tried to prove it, but I dont know the construction of $qR$. In fact,

$qR=q left( mathbbZ[X]/(X^n+1)right)$; is it equal to $(qmathbbZ)[X]/(X^n+1)$?? because $X^n+1 notin qmathbbZ[X]$. Please if someoene can understand this things, and prove the isomorphism (ring isomorphism).

abstract-algebra ring-theory ideals quotient-spaces ring-isomorphism

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asked Jul 21 at 17:05

C.S.

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up vote
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favorite

Let $R= mathbbZ[X]/(X^n+1)$ for some sufficiently large $n$. For $q geq 2$, I want to show that $R/qR cong mathbbZ_q[X]/(X^n+1)$.

I've tried to prove it, but I dont know the construction of $qR$. In fact,

$qR=q left( mathbbZ[X]/(X^n+1)right)$; is it equal to $(qmathbbZ)[X]/(X^n+1)$?? because $X^n+1 notin qmathbbZ[X]$. Please if someoene can understand this things, and prove the isomorphism (ring isomorphism).

abstract-algebra ring-theory ideals quotient-spaces ring-isomorphism

share|cite|improve this question

asked Jul 21 at 17:05

C.S.

134

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $R= mathbbZ[X]/(X^n+1)$ for some sufficiently large $n$. For $q geq 2$, I want to show that $R/qR cong mathbbZ_q[X]/(X^n+1)$.

I've tried to prove it, but I dont know the construction of $qR$. In fact,

$qR=q left( mathbbZ[X]/(X^n+1)right)$; is it equal to $(qmathbbZ)[X]/(X^n+1)$?? because $X^n+1 notin qmathbbZ[X]$. Please if someoene can understand this things, and prove the isomorphism (ring isomorphism).

abstract-algebra ring-theory ideals quotient-spaces ring-isomorphism

share|cite|improve this question

asked Jul 21 at 17:05

C.S.

134

Let $R= mathbbZ[X]/(X^n+1)$ for some sufficiently large $n$. For $q geq 2$, I want to show that $R/qR cong mathbbZ_q[X]/(X^n+1)$.

I've tried to prove it, but I dont know the construction of $qR$. In fact,

$qR=q left( mathbbZ[X]/(X^n+1)right)$; is it equal to $(qmathbbZ)[X]/(X^n+1)$?? because $X^n+1 notin qmathbbZ[X]$. Please if someoene can understand this things, and prove the isomorphism (ring isomorphism).

abstract-algebra ring-theory ideals quotient-spaces ring-isomorphism

share|cite|improve this question

asked Jul 21 at 17:05

C.S.

134

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 17:05

C.S.

134

asked Jul 21 at 17:05

C.S.

134

asked Jul 21 at 17:05

C.S.

134

134

add a comment | 

add a comment | 

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Consider the following exact sequence of abelian group:
$$0to(X^n+1)Bbb Z[X]toBbb Z[X]to Rto0$$
By tensoring with $Bbb Z/qBbb Z$ we get the following exact sequence:
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)otimesBbb Z[X]to(Bbb Z/qBbb Z)otimes Rto0$$
Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)[X]to R/qRto0$$
Since the image of $(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]$ into $(Bbb Z/qBbb Z)[X]$ is the ideal $(X^2+1)(Bbb Z/qBbb Z)[X]$, we get the required isomorphism
$$R/qRcong(Bbb Z/qBbb Z)[X]/(X^2+1)(Bbb Z/qBbb Z)[X]$$

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edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

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1 Answer
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active

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1 Answer
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active

oldest

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active

oldest

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up vote
0
down vote

Consider the following exact sequence of abelian group:
$$0to(X^n+1)Bbb Z[X]toBbb Z[X]to Rto0$$
By tensoring with $Bbb Z/qBbb Z$ we get the following exact sequence:
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)otimesBbb Z[X]to(Bbb Z/qBbb Z)otimes Rto0$$
Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)[X]to R/qRto0$$
Since the image of $(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]$ into $(Bbb Z/qBbb Z)[X]$ is the ideal $(X^2+1)(Bbb Z/qBbb Z)[X]$, we get the required isomorphism
$$R/qRcong(Bbb Z/qBbb Z)[X]/(X^2+1)(Bbb Z/qBbb Z)[X]$$

share|cite|improve this answer

edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

add a comment | 

up vote
0
down vote

Consider the following exact sequence of abelian group:
$$0to(X^n+1)Bbb Z[X]toBbb Z[X]to Rto0$$
By tensoring with $Bbb Z/qBbb Z$ we get the following exact sequence:
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)otimesBbb Z[X]to(Bbb Z/qBbb Z)otimes Rto0$$
Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)[X]to R/qRto0$$
Since the image of $(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]$ into $(Bbb Z/qBbb Z)[X]$ is the ideal $(X^2+1)(Bbb Z/qBbb Z)[X]$, we get the required isomorphism
$$R/qRcong(Bbb Z/qBbb Z)[X]/(X^2+1)(Bbb Z/qBbb Z)[X]$$

share|cite|improve this answer

edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

add a comment | 

up vote
0
down vote

up vote
0
down vote

Consider the following exact sequence of abelian group:
$$0to(X^n+1)Bbb Z[X]toBbb Z[X]to Rto0$$
By tensoring with $Bbb Z/qBbb Z$ we get the following exact sequence:
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)otimesBbb Z[X]to(Bbb Z/qBbb Z)otimes Rto0$$
Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)[X]to R/qRto0$$
Since the image of $(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]$ into $(Bbb Z/qBbb Z)[X]$ is the ideal $(X^2+1)(Bbb Z/qBbb Z)[X]$, we get the required isomorphism
$$R/qRcong(Bbb Z/qBbb Z)[X]/(X^2+1)(Bbb Z/qBbb Z)[X]$$

share|cite|improve this answer

edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

Consider the following exact sequence of abelian group:
$$0to(X^n+1)Bbb Z[X]toBbb Z[X]to Rto0$$
By tensoring with $Bbb Z/qBbb Z$ we get the following exact sequence:
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)otimesBbb Z[X]to(Bbb Z/qBbb Z)otimes Rto0$$
Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence
$$(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]to(Bbb Z/qBbb Z)[X]to R/qRto0$$
Since the image of $(Bbb Z/qBbb Z)otimes(X^n+1)Bbb Z[X]$ into $(Bbb Z/qBbb Z)[X]$ is the ideal $(X^2+1)(Bbb Z/qBbb Z)[X]$, we get the required isomorphism
$$R/qRcong(Bbb Z/qBbb Z)[X]/(X^2+1)(Bbb Z/qBbb Z)[X]$$

share|cite|improve this answer

edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 19:57

edited Jul 21 at 19:57

edited Jul 21 at 19:57

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

answered Jul 21 at 19:48

Fabio Lucchini

5,62411025

5,62411025

add a comment | 

add a comment | 

 

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